CHEM 160 General Chemistry II Lecture Presentation Chemical Thermodynamics Chapter 19 November 19, 2018 Chapter 19

Thermodynamics Thermodynamics study of energy and its transformations heat and energy flow Helps determine natural direction of reactions Allows us to predict if a chemical process will occur under a given set of conditions Organized around three fundamental laws of nature 1st law of thermodynamics 2nd law of thermodynamics 3rd law of thermodynamics November 19, 2018 Chapter 19

First Law of Thermodynamics Energy can be neither created nor destroyed Energy of the universe is constant Important concepts from thermochemistry Enthalpy Hess’s law Purpose of 1st Law Energy bookkeeping How much energy? Exothermic or endothernic? What type of energy? November 19, 2018 Chapter 19

Spontaneous vs Nonspontaneous Spontaneous process Occurs without outside assistance in the form of energy (occurs naturally) Product-favored at equilibrium May be fast or slow May be influenced by temperature Nonspontaneous process Does not occur without outside assistance Reactant-favored at equilibrium All processes which are spontaneous in one direction cannot be spontaneous in the reverse direction Spontaneous processes have a definite direction Spontaneous processes are irreversible November 19, 2018 Chapter 19

Spontaneous vs. Nonspontaneous Spontaneous Processes Gases expand into larger volumes H2O(s) melts above 0C H2O(l) freezes below 0C NH4NO3 dissolves spontaneously in H2O Steel (iron) rusts in presence of O2 and H2O Wood burns to form CO2 and H2O CH4 gas burns to form CO2 and H2O Reverse processes are not spontaneous! nonspontaneous November 19, 2018 Chapter 19

Spontaneous vs. Nonspontaneous Spontaneous Processes H2O(s) melts above 0C (endothermic) NH4NO3 dissolves spontaneously in H2O (endothermic) Steel (iron) rusts in presence of O2 and H2O (exothermic) Wood burns to form CO2 and H2O (exothermic) CH4 gas burns to form CO2 and H2O (exothermic) Heat change alone is not enough to predict spontaneity because energy is conserved Many, but not all, spontaneous processes are exothermic November 19, 2018 Chapter 19

Factors That Favor Spontaneity Two thermodynamic properties of a system are considered when determining spontaneity: Enthalpy, H Many, but not all, spontaneous processes tend to be exothermic as already noted Entropy, S (J/K) Measure of the disorder of a system Many, but not all, spontaneous processes tend to increase disorder of the system November 19, 2018 Chapter 19

Entropy Entropy, S (J/K) describes # of ways the particles in a system can be arranged in a given state More arrangements = greater entropy S = heat change/T = q/T S = Sfinal - Sinitial  S > 0 represents increased randomness or disorder November 19, 2018 Chapter 19

Patterns of Entropy Change For the same or similar substances: Ssolid < Sliquid < Sgas solid vapor November 19, 2018 Chapter 19 liquid

Patterns of Entropy Change Particles farther apart, occupy larger volume of space; even more positions available to particles Rigidly held particles; few positions available to particles Particles free to flow; more positions available for particles solid vapor November 19, 2018 Chapter 19 liquid

Patterns of Entropy Change least ordered less ordered most ordered solid vapor November 19, 2018 Chapter 19 liquid

Patterns of Entropy Change Solution formation usually leads to increased entropy particles more disordered solvent solute solution November 19, 2018 Chapter 19

Patterns of Entropy Change Chemical Reactions If more gas molecules produced than consumed, S increases. (Srxn > 0) If only solids, ions and/or liquids involved, S increases if total # particles increases. November 19, 2018 Chapter 19

fewer arrangements possible so lower entropy more arrangements possible so higher entropy November 19, 2018 Chapter 19

Patterns of Entropy Change Increasing temperature increases entropy System at T1 (S1) System at T2 (T2 > T1) (S2 > S1) November 19, 2018 Chapter 19

Patterns of Entropy Change more energetic molecular motions less energetic molecular motions System at T1 (S1) System at T2 (T2 > T1) (S2 > S1) November 19, 2018 Chapter 19

Third Law of Thermodynamics If increasing T increases S, then the opposite should be true also. Is it possible to decrease T to the point that S is zero? At what T does S = 0? If entropy is zero, what does that mean? November 19, 2018 Chapter 19

Third Law of Thermodynamics Entropy of a perfect crystalline substance at 0 K is zero No entropy = highest order possible Why? Purpose of 3rd Law Allows S to be measured for substances S = 0 at 0 K S = heat change/temperature = q/T S = standard molar entropy November 19, 2018 Chapter 19

Standard entropy, S° (J/K) 50 40 Standard entropy, S° (J/K) 30 20 10 50 100 150 200 250 300 November 19, 2018 Temperature (K) Chapter 19

Gas Liquid Solid Standard entropy, S° (J/K) Temperature (K) 50 40 30 20 Solid 10 50 100 150 200 250 300 November 19, 2018 Temperature (K) Chapter 19

Standard Molar Entropies of Selected Substances at 298 K S (J/K) H2O(l) 69.9 NaCl(s) 72.3 H2O(g) 188.8 NaCl(aq) 115.5 I2(s) 116.7 Na2CO3(s) 136.0 I2(g) 260.6 CH4(g) 186.3 Na(s) 51.5 C2H6(g) 229.5 K(s) 64.7 C3H8(g) 269.9 Cs(s) 85.2 C4H10(g) 310.0 November 19, 2018 Chapter 19

Entropy versus Probability Systems tend to move spontaneously towards increased entropy. Why? Entropy is related to probability Disordered states are more probable than ordered states S = k(lnW) k (Boltzman’s constant) = 1.38 x 10-23 J/K W = # possible arrangements in system November 19, 2018 Chapter 19

Consider why gases tend to isothermally expand into larger volumes. Isothermal Gas Expansion Consider why gases tend to isothermally expand into larger volumes. Gas Container = two bulbed flask Ordered State Gas Molecules November 19, 2018 Chapter 19

Isothermal Gas Expansion Gas Container Ordered State S = k(ln 1) = (1.38 x 10-23 J/K)(0) = 0 J/K November 19, 2018 Chapter 19

November 19, 2018 Chapter 19

November 19, 2018 Chapter 19

November 19, 2018 Chapter 19

November 19, 2018 Chapter 19

November 19, 2018 Chapter 19

Disordered States November 19, 2018 Chapter 19

Disordered States November 19, 2018 Chapter 19

Disordered States More probable that the gas molecules will disperse between two halves than remain on one side November 19, 2018 Chapter 19

Disordered States Driving force for expansion is entropy (probability); gas molecules have a tendency to spread out November 19, 2018 Chapter 19

Disordered States S = k(ln 7) = (1.38 x 10-23 J/K)(1.95) = 2.7 x 10-23 J/K November 19, 2018 Chapter 19

Total Arrangements Stotal = k(ln 23) = k(ln 8) = (1.38 x 10-23 J/K)(1.79) = 2.9 x 10-23 J/K November 19, 2018 Chapter 19

2nd Law of Thermodynamics The entropy of the universe increases in any spontaneous process (Suniv > 0) Increased disorder in the universe is the driving force for spontaneity November 19, 2018 Chapter 19

2nd Law of Thermodynamics Suniv = Ssysyem + Ssurroundings Suniv > 0, process is spontaneous Suniv = 0, process at equilibrium Suniv < 0, process is nonspontaneous Process is spontaneous in reverse direction November 19, 2018 Chapter 19

2nd Law of Thermodynamics Spontaneous Processes H2O(s) melts above 0C (endothermic) Steel (iron) rusts in presence of O2 and H2O (exothermic) 4Fe(s) + 3O2(g)  2Fe2O3(s) CH4 gas burns to form CO2 and H2O (exothermic) CH4(g) + 2O2(g)  CO2(g) + 2H2O(g) Each process increases Suniverse. November 19, 2018 Chapter 19

2nd Law of Thermodynamics To determine Suniv for a process, both Ssystem and Ssurroundings need to be known: Ssysyem related to matter dispersal in system Ssurroundings determined by heat exchange between system and surroundings and T at which it occurs Sign of Ssurroundings depends on whether process in system is endothermic or exothermic Why? Magnitude of Ssurroundings depends on T Ssurroundings = -Hsystem/T November 19, 2018 Chapter 19

Entropy Changes in a System For any reaction: S°rxn = nS°(products) - mS°(reactants) Where n and m are stoichiometric coefficients November 19, 2018 Chapter 19

Example 1 (1 on Example Problem Handout) Using standard molar entropies, calculate S°rxn for the following reaction at 25°C: 2SO2(g) + O2(g) --> 2SO3(g) S° = 248.1 205.1 256.6 (J · K-1mol-1) (Ans.: -187.9 J/K) November 19, 2018 Chapter 19

Gibbs Free Energy, G 2nd law: Suniv = Ssys + Ssurr = Ssys - Hsys/T Rearrange (multiply by -T) -TSuniv = -TSsys + Hsys = Hsys - TSsys -TSuniv = G (-TSuniv = G) G = Gibbs free energy (J or kJ) G = H - TS and G = Hsys - TSsys G° = H°sys - TS°sy (if at standard state) November 19, 2018 Chapter 19

Gibbs Free Energy Summary of Conditions for Spontaneity G < 0 November 19, 2018 Chapter 19

Gibbs Free Energy Summary of Conditions for Spontaneity G < 0 reaction is spontaneous in the forward direction (Suniv > 0) G > 0 G = 0 November 19, 2018 Chapter 19

Gibbs Free Energy Summary of Conditions for Spontaneity G < 0 reaction is spontaneous in the forward direction (Suniv > 0) G > 0 reaction is nonspontaneous in the forward direction (Suniv < 0) G = 0 November 19, 2018 Chapter 19

Gibbs Free Energy Summary of Conditions for Spontaneity G < 0 reaction is spontaneous in the forward direction (Suniv > 0) G > 0 reaction is nonspontaneous in the forward direction (Suniv < 0) G = 0 system is at equilibrium (Suniv = 0) November 19, 2018 Chapter 19

Example 2 (2 in Example Problem Handout) For a particular reaction, Hrxn = 53 kJ and Srxn = 115 J/K. Is this process spontaneous a) at 25°C, and b) at 250°C? (c) At what temperature does Grxn = 0? (ans.: a) G = 18.7 kJ, nonspontaneous; b) –7.1 kJ, spontaneous; c) 460.9 K or 188C) November 19, 2018 Chapter 19

Gibbs Free Energy and Temperature G = H - TS S > 0 H < 0 Spontaneous at all T S < 0 H > 0 Not spontaneous at any T S > 0 H > 0 Spontaneous at high T; nonspontaneous at low T S < 0 H < 0 Spontaneous at low T; nonspontaneous at high T November 19, 2018 Chapter 19

Gibbs Free Energy and Temperature Spontaneous Processes H2O(s) melts above 0C (endothermic) H > 0, S > 0 Steel (iron) rusts in presence of O2 and H2O at 25 C (exothermic) 4Fe(s) + 3O2(g)  2Fe2O3(s) H < 0, S < 0 G < 0 for each process T determines sign of G: G = H -TS November 19, 2018 Chapter 19

Calculating Free Energy Changes For any reaction: G°rxn = Gf°(products) - mGf°(reactants) Where n and m are stoichiometric coefficients and Gf° = standard free energy of formation November 19, 2018 Chapter 19

Example 3 (3 on Example Problem Handout) Calculate the standard free energy change for the reaction at 25°C: CH4(g) + 2O2(g)  CO2(g) + 2H2O(l) G°f = -50.8 0 -394.4 -237.4 (kJ/mol) November 19, 2018 Chapter 19

Gibbs Free Energy What does Gibbs free energy represent? For a spontaneous process: Maximum amount of energy released by the system that can do useful work on the surroundings Energy available from spontaneous process that can be used to drive nonspontaneous process For a nonspontaneous process: Minimum amount of work that must be done to force the process to occur In actuality, Gspont > Gnonspont November 19, 2018 Chapter 19

Gibbs Free Energy Conversion of rust to iron 2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS) To convert iron to rust, G must be provided from spontaneous rxn 6CO + 3O2  6CO2 G = -1543 kJ (S) November 19, 2018 Chapter 19

Gibbs Free Energy Conversion of rust to iron 2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS) To convert iron to rust, G must be provided from spontaneous rxn 2Fe2O3  4Fe + 3O2 G = 1487 kJ (NS) 6CO + 3O2  6CO2 G = -1543 kJ (S) 2Fe2O3 + 6CO  4Fe + 6CO2  G = -56 kJ (S) Reactions are “coupled” November 19, 2018 Chapter 19

Gibbs Free Energy Many biological rxns essential for life are NS Spontaeous rxns used to “drive” the NS biological rxns Example: photosynthesis 6CO2 + 6H2O  C6H12O6 + 6O2  G > 0 What spontaneous rxns drive photosynthesis? November 19, 2018 Chapter 19

proteins, cells etc., lower S amino acids, sugars, etc., higher S sun big ball of G = G released high free energy DGsurr < 0 proteins, cells etc., lower S C6H12O6, O2 ATP NS Sp NS Sp NS CO2, H2O amino acids, sugars, etc., higher S ADP photosynthesis solar nuclear reactions low free energy November 19, 2018 Chapter 19

Free Energy and Equilibrium G (Nonstandard State) G = G° + RTlnQ R = gas constant (8.314 J/K·mol) T = temperature (K) Q = reaction quotient November 19, 2018 Chapter 19

Example 4 (4 on Example Problem Handout) Calculate Grxn for the reaction below: 2A(aq) + B(aq)  C(aq) + D(g) if G°rxn = 9.9 x 103 J/mol and (a) [A] = 0.8 M, [B] = 0.5 M, [C] = 0.05 M, and PD = 0.05 atm, and (b) [A] = 0.1 M, [B] = 1 M, [C] = 0.5 M, and PD = 0.5 atm. Is the reaction spontaneous under these conditions? (ans.: a) –2121 J, spontaneous; b) 17875 J, nonspontaneous) November 19, 2018 Chapter 19

Free Energy and Equilibrium At equilibrium: Grxn = 0 and Q = K 0 = G°rxn + RTlnK G°rxn = -RTlnK November 19, 2018 Chapter 19

Example 5 (5 on Example Problem Handout) Calculate G°rxn for the ionization of acetic acid, HC2H3O2 (Ka = 1.8 x 10-5) at 25°C. Is this reaction spontaneous under standard state conditions? (ans.: 27 kJ) November 19, 2018 Chapter 19

Example 6 (6 on Example Problem Handout) Calculate G° for the neutralization of a strong acid with a strong base at 25°C. Is this process spontaneous under these conditions? For the reaction below, K = 1.0 x 1014. H+ + OH-  H2O (ans.: -80 kJ) November 19, 2018 Chapter 19

Example 7 Calculate Keq for a reaction at (a) 25°C and (b) 250°C if H°rxn = 42.0 kJ and S°rxn = 125 J/K. At which temperature is this process product favored? (ans.: a) 0.15; b) 216; 250C) November 19, 2018 Chapter 19

DGrxn < 0 Spontaneous Reaction equilibrium position Pure reactants Pure products DGrxn < 0 Gproducts Greactants extent of reaction November 19, 2018 Chapter 19

Spontaneous Reaction At equilibrium, any change requires an uphill climb in energy, DGrxn > 0 Gproducts Greactants Pure reactants Pure products extent of reaction November 19, 2018 Chapter 19

Calculating G for Processes G may be calculated in one of several ways depending on the information known about the process of interest G = H - TS; G° = H° - TS° G°rxn = Gf°(products) - mGf°(reactants) G = G° + RTlnQ G°rxn = -RTlnK November 19, 2018 Chapter 19

Entropy and Life Processes If the 2nd law is valid, how is the existence of highly-ordered, sophisticated life forms possible? growth of a complex life form represents an increase in order (less randomness) lower entropy November 19, 2018 Chapter 19

Entropy and Life Processes Organisms “pay” for their increased order by increasing Ssurr. Over lifetime, Suniv > 0. CO2 H2O heat November 19, 2018 Chapter 19